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GATE 2020Mechanical Engineering
(Volume - I)
TOPIC WISE GATE SOLUTIONS
1987 - 2019
TM
GATE A PCADEMY UBLICATIONS®
S. No. Topics Page No.
1. Fluid Mechanics 1.1 - 1.000
1. Properties of Fluid 1.1 - 1.00
2. Pressure & Its Measurement 1.00 - 1.00
3. Hydrostatic Forces 1.00 - 1.00
4. Buoyancy & Floatation 1.00 - 1.00
5. Kinematics of Fluid 1.00 - 1.00
6. Dynamics of Fluid 1.00 - 1.00
7. Laminar & Turbulent Flow, Viscous Flow, Flow Through Pipes 1.00 - 1.00
8. Boundary Layer Theory 1.00 - 1.00
9. Hydraulic Machines 1.00 - 1.00
2. Thermodynamics 2.1 - 2.000
1. Thermodynamics System & Process 2.1 - 2.00
2. First Law of Thermodynamics 2.00 - 2.00
3. Second Law of Thermodynamics 2.00 - 2.00
4. Entropy 2.00 - 2.00
5. Availability & Irreversibility 2.00 - 2.00
6. Pure Substances 2.00 - 2.00
7. Steam Power Cycles 2.00 - 2.00
8. Gas Power Cycles 2.00 - 2.00
9. Internal Combustion Engine 2.00 - 2.00
3. Heat Transfer 3.1 - 3.000
1. Conduction 3.1 - 3.0
2. Fins & Unsteady Heat Transfer 3.0 - 3.00
3. Free & Forced Convection 3.00 - 3.00
4. Heat Exchanger 3.00 - 3.00
5. Radiation 3.00 - 3.00
CONTENTS
4. Refrigeration & Air-Conditioning 4.1 - 4.000
1. Vapour Refrigeration 4.1 - 4.0
2. Properties of Moist Air & Psychrometric Processes 4.0 - 4.00
3. Heat Pumps & Cycles 4.00- 4.00
5. Mechanics of Solids 5.1 - 5.000
1. Stress & Strain 5.1 - 5.0
2. Complex Stress 5.0 - 5.00
3. Elastic Constants & Theory of Failure 5.00 - 5.00
4. Thin Cylinder 5.00 - 5.00
5. Shear Force & Bending Moment Diagrams 5.00 - 5.00
6. Bending of Beam 5.00 - 5.00
7. Torsion of Shaft 5.00 - 5.00
8. Springs 5.00 - 5.00
9. Euler's Theory of Column 5.00 - 5.00
10. Strain Energy & Thermal Stresses 5.00 - 5.00
6. Machine Design 6.1 - 6.000
1. Design Against Static Load (Theory of Failure) 6.1 - 6.0
2. Design Against Dynamic Load (Fatigue Strength & SN Diagram) 6.0 - 6.00
3. Gears 6.00 - 6.00
4. Bearings, Shaft & Keys 6.00 - 6.00
5. Clutches, Ropes & Belts 6.00 - 6.00
6. Brakes 6.00 - 6.00
7. Joints (Bolted, Riveted & Welded) 6.00 - 6.00
8. Power Screws & Springs 6.00 - 6.00
7. General Aptitude 7.1 - 7.000
1. Numerical Ability 7.1 – 7.0
2. Logical Reasoning 7.0 - 7.00
3. Verbal Ability 7.00 - 7.00
© Copyrightwww.gateacademy.co.inH Oead ffice : A/114-115, Smriti Nagar, Bhilai (C.G.), Contact : 9713113156, 9589894176
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1.1 Thermal conductivity is lower for
[1 Mark] (A) Wood (B) Air (C) Water at 0100 C (D) Steam at 1 bar
1.2 Match the property with their units
[2 Marks] Property Units
A. Bulk modulus 1. W/s B. Thermal
conductivity 2. 2N/m
C. Heat transfer coefficient
3. 3N/m
D. Heat flow rate 4. W 5. W/mK
6. 2W/m K
1.3 For a current carrying wire of 20 mm
diameter exposed to air 2( 25W/m K)h , maximum heat distribution occurs when the thickness of insulation is ( 0.5W/mK)k [2 Marks]
(A) 20 mm (B) 10 mm (C) 25 mm (D) 0 mm
1.4 Two insulating materials of thermal
conductivity k and 2k are available for lagging a pipe carrying a hot fluid. If the radial thickness of each material is same.
[1 Mark] (A) Material with higher thermal
conductivity should be used for inner layer and one with lower thermal conductivity for the outer.
(B) Material with lower thermal conductivity should be used for inner layer and one with higher thermal conductivity for the outer.
(C) It is immaterial in which sequence the insulating material are used.
(D) It is not possible to judge unless numerical values of dimensions are given.
1.5 For a given heat flow and for the same
thickness, the temperature drop across the material will be maximum for
[1 Mark] (A) Copper. (B) Steel. (C) Glass wool. (D) Refractory brick.
1 Conduction
1990 IISc Bangalore
1991 IIT Madras
1993 IIT Bombay
1994 IIT Kharagpur
1996 IISc Bangalore
3.2 Topic Wise GATE Solutions [ME] GATE ACADEMY®
© Copyrightwww.gateacademy.co.inH Oead ffice : A/114-115, Smriti Nagar, Bhilai (C.G.), Contact : 9713113156, 9589894176
B Oranch ffice : Raipur : 79743-90037, Bhopal : 83198-88401
1.6 The temperature variation under steady
heat conduction across a composite slab of two materials with thermal conductivities 1k and 2k is shown in figure then, which one of the following statements holds? [2 Marks]
(A) 1 2k k (B) 1 2k k
(C) 1 0k (D) 1 2k k
1.7 It is proposed to coat a 1 mm diameter
wire with enamel paint ( 0.1W/mK)k to increase heat transfer with air. If the air side heat transfer co-efficient is
2100 W/m K , the optimum thickness of enamel paint should be [2 Marks]
(A) 0.25 mm (B) 0.5 mm
(C) 1 mm (D) 2 mm
1.8 In descending order of magnitude, the
thermal conductivity of (a) pure iron, (b) liquid water, (c) Saturated water vapour, (d) Pure aluminum can be arranged as
[1 Mark]
(A) a b c d (B) b c a d
(C) d a b c (D) d c b a
Heat is being transferred by convection from water at 048 C to a glass plate whose surface that is exposed to the water is at 040 C . The thermal conductivity of water is 0.6 W/mK and the thermal conductivity of glass is 1.2 W/mK. The spatial gradient of temperature in the water at the water glass interface is
41 10dT
dy K/m.
1.9 The value of the temperature gradient in
the glass at the water glass interface in K/m is [2 Marks]
(A) 42 10 (B) 0.0
(C) 40.5 10 (D) 42 10 1.10 The heat transfer coefficient h in
2W/m K is [2 Marks] (A) 0.0 (B) 4.8 (C) 6 (D) 750
1.11 One dimensional unsteady state heat
transfer equation for a sphere with heat generation at the rate of ‘q’ can be written [1 Mark]
(A) 1 1T q Tr
r r r k t
(B) 22
1 1T q Tr
r r r k t
1k
1T2k
3T
2T
040 C
048 C
y
WaterGlass
1998 IIT Delhi
1999 IIT Bombay
2001 IIT Kanpur
2003 IIT Madras
Common Data for Questions 1.9 & 1.10
2004 IIT Delhi
3.3GATE ACADEMY® Heat Transfer : Conduction
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(C) 2
2
1T q T
r k t
(D) 2
2
1( ) q TrT
r k t
1.12 A stainless steel tube ( 19 W/mK)sk of 2 cm ID and 5 cm OD in insulated with 3 cm thick asbestos ( 0.2 W/mK)ak . If the temperature difference between the inner most and outermost surface is
0600 C , the heat transfer rate per unit length is [2 Marks]
(A) 0.94 W/m (B) 9.44 W/m
(C) 944.72 W/m (D) 9447.21 W/m
1.13 A well machined steel plate of thickness
L is kept such that the wall temperatures are hT and cT as shown in the figure below. A smooth copper plate of the same thickness L is now attached to the steel plate without any gap as indicated in the figure below. The temperature at the interface is iT . The temperatures of
the outer walls are still the same at hT
and cT . The heat transfer rates are 1q and
2q per unit area in the two cases respectively in the direction shown.
Which of the following statements is correct? [1 Mark]
(A) h i cT T T and 1 2q q
(B) h i cT T T and 1 2q q
(C) ( ) / 2h i cT T T and 1 2q q
(D) ( ) / 2i h cT T T and 1 2q q
1.14 In a case of one dimensional heat conduction in a medium with constant properties, T is the temperature at
position x, at time t. Then T
t
is
proportional to [1 Mark]
(A) Tx
(B) T
t
(C) 2T
x t
(D) 2
2
T
x
1.15 Heat flows through a composite slab, as shown below. The depth of the slab is 1 m. The k values are in W/mK. The overall thermal resistance in K/W is
[2 Marks]
(A) 17.2 (B) 21.9
(C) 28.6 (D) 39.2
1q1q
hT
L
CT
STEEL
Ti
2q 2q
cThT
L
CopperSteel
L
1 0.02k �
0.5m
2 0.10k �
3 0.04k �
0.25m
3
2
1
0.5
m
1 mq
2005 IIT Bombay
3.4 Topic Wise GATE Solutions [ME] GATE ACADEMY®
© Copyrightwww.gateacademy.co.inH Oead ffice : A/114-115, Smriti Nagar, Bhilai (C.G.), Contact : 9713113156, 9589894176
B Oranch ffice : Raipur : 79743-90037, Bhopal : 83198-88401
1.16 In a composite slab, the temperature at
the interface inter( )T between two materials is equal to the average of the temperature at the two ends. Assuming steady one dimensional heat conduction, which of the following statements is true about the respective thermal conducti-vities? [1 Mark]
(A) 1 22k k (B) 1 2k k
(C) 1 22 3k k (D) 1 22k k 1.17 With an increase in the thickness of
insulation around a circular pipe, heat loss to surrounding due to [2 Marks]
(A) Convection increases while heat due to conduction decreases.
(B) Convection decreases, while that due to conduction increases.
(C) Convection and conduction decreases.
(D) Convection and conduction increases.
1.18 A long glass cylinder of inner diameter
0.03 m and outer diameter 0.05 m carries hot fluid inside. If the thermal conductivity of glass 1.05W/mKk , the thermal resistance (K/W) per unit length of the cylinder is [2 Marks]
(A) 0.031 (B) 0.077 (C) 0.17 (D) 0.34
1.19 Heat is being transferred convectively from a cylindrical nuclear reactor fuel rod of 50 mm diameter to water at 075 Cunder steady state condition, the rate of heat generation within the fuel element is 6 350 10 W/m and the convective heat
transfer coefficient is 21 kW/m K , the outer surface temperature of the fuel element would be [2 Marks]
(A) 0700 C (B) 0625 C (C) 0500 C (D) 0400 C
Consider steady one dimensional heat flow in a plate of 20 mm thickness with a uniform heat generation of 380MW/m . The left and right
faces are kept at constant temperature of 0160 Cand 0120 C respectively. The plate has a constant thermal conductivity of 200 W/mK. 1.20 The location of maximum temperature
within the plate from its face is [2 Marks]
(A) 15 mm (B) 10 mm (C) 5 mm (D) 0 mm 1.21 The maximum temperature within the
plate in 0 C is [2 Marks] (A) 160 (B) 165 (C) 200 (D) 250
1.22 Steady two dimensional heat conduction
takes place in the body shown in the figure below. The normal temperature gradients at surface P and Q can be considered to be uniform. The
temperature gradients T
x
over a surface
bb
Ti
1k
1T
2T
2k
2 b2 b
2006 IIT Kharagpur
2007 IIT Kanpur
Common Data for Questions 1.20 & 1.21
2008 IISc Bangalore
3.5GATE ACADEMY® Heat Transfer : Conduction
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Q is equal to 10 K/m. Surfaces P and Q are maintained at constant temperature as shown in the figure, while the remaining part of the boundary is insulated.
The body has a constant thermal conductivity of 0.1 W/mK. The value of
T
y
and T
x
at surface P are [2 Marks]
(A) 20K/m, 0K/mT T
x y
(B) 0K/m, 10K/mT T
x y
(C) 10K/m, 10K/mT T
x y
(D) 0K/m, 20K/mT T
x y
1.23 Consider one-dimensional steady state
heat conduction, without heat generation, in a plane wall, with boundary conditions as shown in the figure below. The conductivity of the wall is given by 0k k bT , where 0k
and b are positive constants, and T is temperature.
As x increases, the temperature gradient dT
dx
will [1 Mark]
(A) Remain constant. (B) Be zero. (C) Increase. (D) Decrease. 1.24 Consider one-dimensional steady state
heat conduction along x-axis ( 0 x L ), through a plane with the boundary surfaces ( 0x and x L ) maintained at temperatures of 00 C and 0100 C . Heat is generated uniformly throughout the wall. Choose the correct statements.
[1 Mark] (A) The direction of heat transfer will be
from the surface at 0100 C to the surface at 00 C .
(B) The maximum temperature inside the wall must be greater than 0100 C.
(C) The temperature distribution is linear within the wall.
(D) The temperature distribution is symmetric about the mid-plane of the wall.
1.25 Consider a long cylindrical tube of inner
and outer radii, ir and 0r respectively, length L and thermal conductivity k. its inner and outer surfaces are maintained at 1T and 0T , respectively 1 0( )T T . Assuming one-dimensional steady state heat conduction in the radial direction, the thermal resistance in the wall of the tube is [1 Mark]
(A) 0
1 ln2
ir
kL r
(B) 2 i
L
rk
(C) 01 ln2 i
r
kL r
(D) 01 ln4 i
r
k L r
Surface
Surface
0,0 CQ
0,100 CP
2 m
1m
y
x
T1 T2 Where
x
2 1T T�
2013 IIT Bombay
2014 IIT Kharagpur
3.6 Topic Wise GATE Solutions [ME] GATE ACADEMY®
© Copyrightwww.gateacademy.co.inH Oead ffice : A/114-115, Smriti Nagar, Bhilai (C.G.), Contact : 9713113156, 9589894176
B Oranch ffice : Raipur : 79743-90037, Bhopal : 83198-88401
1.26 As the temperature increases, the thermal conductivity of a gas [1 Mark]
(A) Increases (B) Decreases (C) Remain constant (D) Increases up to a certain temperature
and then decreases 1.27 Consider one dimensional steady state
heat conduction across a wall (as shown in figure below) of thickness 30 mm and thermal conductivity 15 W/mK. At
0x a constant heat flux, 5 2'' 1 10 W/mq is applied. On the
other side of the wall, heat is removed from the wall by convection with a fluid at 025 C and heat transfer coefficient of
2250 W/m K. The temperature (in 0 C ), at 0x is ________. [2 Marks]
1.28 A material P of thickness 1 mm is
sandwiched between two steel slabs, as shown in the figure below. A heat flux
210kW/m is supplied to one of the steel slabs as shown. The boundary temperatures of the slabs are indicated in the figure. Assume thermal conductivity of this steel is 10 W/mK. considering one-dimensional steady state heat conduction for the configuration, the thermal conductivity (k, in W/mK) of material P is ________. [2 Marks]
1.29 Heat transfer through a composite wall
is shown in figure. Both the sections of the wall have equal thickness ( l ). The conductivity of one section is k and that of the other is 2k. The left face of the wall is at 600 K and the right face is at 300 K. [2 Marks]
The interface temperature (in K)iT of
the composite wall is ______. 1.30 A plane wall has a thermal conductivity
of 1.15 W/mK. If the inner surface is at 01100 Cand the outer surface is at
0350 C , then the design thickness (in meter) of the wall to maintain a steady heat flux of 22500 W/m should be ____.
[2 Marks]
1.31 If a foam insulation is added to a 4 cm
outer diameter pipe as shown in the figure, the critical radius of insulation (in cm) is _______. [1 Mark]
0�x
x
025 C� �T
2T
1T
5 2" 1 10 W/m� �q
20 mm 20 mm1 mm
2 360K�T2" 10 kW/m�q
1 500 K�T
?�k
P
Steelslab
Steelslab
l l
300 K600 K iT
k 2k
Heat flow
2015 IIT Kanpur
3.7GATE ACADEMY® Heat Transfer : Conduction
© Copyrightwww.gateacademy.co.inH Oead ffice : A/114-115, Smriti Nagar, Bhilai (C.G.), Contact : 9713113156, 9589894176
B Oranch ffice : Raipur : 79743-90037, Bhopal : 83198-88401
1.32 A 10 mm diameter electrical conductor
is covered by an insulation of 2 mm thickness. The conductivity of the insulation is 0.08 W/mK and the convection coefficient at the insulation surface is 210W/m K . Addition of further insulation of the same material will
[2 Marks]
(A) Increases heat loss continuously
(B) Decreases heat loss continuously
(C) Increases heat loss to a maximum and then decreases heat loss
(D) Decreases heat loss to a minimum and then increases heat loss
1.33 A brick wall ( 0.9 W/mK)k of thickness 0.18 m separates the warm air in a room from the cold ambient air. On a particular winter day, the outside air temperature is 05 C and the room needs
to be maintained at 027 C . The heat transfer coefficient associated with outside air is 220 W/m K . Neglecting the convective resistance of the air inside the room, the heat loss, in 2W/m , is [2 Marks]
(A) 88 (B) 110
(C) 128 (D) 160
1.34 A cylindrical uranium fuel rod of radius 5 mm in a nuclear reactor is generating heat at the rate of 7 34 10 W/m . The rod is cooled by a liquid (convective heat transfer coefficient 21000 W/m K ) at
025 C . At steady state, the surface temperature (in K) of the rod is
[2 Marks] (A) 308 (B) 398 (C) 418 (D) 448
1.35 A plastic sleeve of outer radius
0 1 mmr covers a wire (radius 0.5 mmr ) carrying electric current.
Thermal conductivity of the plastic is 0.15 W / m K . The heat transfer coefficient on the outer surface of the sleeve exposed to air is 25 2W / m K . Due to the addition of the plastic cover, the heat transfer from the wire to the ambient will [1 Mark]
(A) Increase (B) Remain the same (C) Decrease (D) Be zero 1.36 A hollow cylinder has length ,L inner
radius 1,r outer radius 2 ,r and thermal conductivity k . The thermal resistance of the cylinder for radial conduction is
[1 Mark]
(A)
2
1
ln
2
rr
kL
(B)
1
2
ln
2
rr
kL
(C) 2
1
2
ln
kL
rr
(D) 1
2
2
ln
kL
rr
2
0 2 W/m Kh �
0.1W/mKfoamk �
15 W/m-K�pipek
Pipe
FoamFoam
2016 IISc Bangalore
3.8 Topic Wise GATE Solutions [ME] GATE ACADEMY®
© Copyrightwww.gateacademy.co.inH Oead ffice : A/114-115, Smriti Nagar, Bhilai (C.G.), Contact : 9713113156, 9589894176
B Oranch ffice : Raipur : 79743-90037, Bhopal : 83198-88401
1.37 Steady one-dimensional heat conduction takes place across the faces 1 and 3 of a composite slab consisting of slabs A and B in perfect contact as shown in the figure, where ,A Bk k denote the respective thermal conductivities. Using the data as given in the figure, the interface temperature 2T (in 0 C ) is ______. [1 Mark]
1.38 Heat is generated uniformly in a long
solid cylindrical rod (diameter = 10 mm) at the rate of 7 34 10 W/m . The thermal conductivity of the rod material is 25 W/m K. Under steady state conditions, the temperature difference between the center and the surface of the rod is ________ 0 C [2 Marks]
1.39 A plane slab of thickness L and thermal
conductivity k is heated with a fluid on one side (P) and other side (Q) is maintained at a constant temperature QT
of 025 C , as shown in the figure. The fluid is at 045 C and the surface heat transfer coefficient, h is 210 W/m K . The steady state temperature pT (in 0 C ) of the side which is exposed to the fluid is ______ (correct to two decimal places). [2 Marks]
1.40 A slender rod of length L and diameter
( )d L d and thermal conductivity 1k is joined with another rod of identical dimensions, but of thermal conductivity
2k , to form a composite cylindrical rod of length 2L. The heat transfer in radial direction and contact resistance are negligible. The effective thermal conductivity of the composite rod is
[1 Mark]
(A) 1 2
1 2
2k k
k k (B) 1 2
1 2
k k
k k
(C) 1 2k k (D) 1 2k k
1.41 One-dimensional steady state heat conduction takes place through a solid whose cross section area varies linearly in the direction of heat transfer. Assume there is no heat generation in solid and thermal conductivity of the material is constant and independent of temperature. The temperature distribution in the solid is [1 Mark]
(A) Linearly (B) Quadratic (C) Logarithmic (D) Exponential 1.42 Three slabs are joined together as shown
in the figure. There is no thermal contact resistance at the interfaces. The center slab experiences a non-uniform internal heat generation with an average value equal to 310000 Wm , while the left and right slabs have no internal heat
A B
1 2 3
0
1 130 CT � 0
3 30 CT �20
W/mK
Ak � 100 W/mKBk �
0.1 m 0.3 m
L = 20 cm
TQ = 25 C0h = 10 W/m K
2
0= 45 CT
TP
k = 2.5 W/mK
2017 IIT Roorkee
2018 IIT Guwahati
2019 IIT Madras
3.9GATE ACADEMY® Heat Transfer : Conduction
© Copyrightwww.gateacademy.co.inH Oead ffice : A/114-115, Smriti Nagar, Bhilai (C.G.), Contact : 9713113156, 9589894176
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generation. All slabs have thickness equal to 1 m and thermal conductivity of each slab is equal to 1 15Wm K . The two extreme faces are exposed to fluid with heat transfer coefficient
2 1100 Wm K and bulk temperature 030 C as shown.
The heat transfer in the slabs is assumed to be one dimensional and steady, and all properties are constant. If the left extreme face temperature 1T is measured
to be 0100 C, the right extreme face temperature 2T is _____ 0 C . [2 Marks]
1 m 1 m 1 mLeft extremeface = 100 CT1
0
100 W/m K2
30 C0
100 W/m K2
30 C0
T2
3.10 Topic Wise GATE Solutions [ME] GATE ACADEMY®
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1.1 B 1.2 A-2, B-5, C-6, D-4
1.3 B 1.4 B 1.5 C
1.6 A 1.7 B 1.8 C 1.9 C 1.10 D
1.11 B 1.12 C 1.13 D 1.14 D 1.15 C
1.16 D 1.17 A 1.18 B 1.19 A 1.20 C
1.21 B 1.22 D 1.23 D 1.24 B 1.25 C
1.26 A 1.27 625 1.28 0.1 1.29 400 1.30 0.345
1.31 5 1.32 C 1.33 C 1.34 B 1.35 A
1.36 A 1.37 67.5 1.38 10 1.39 33.88 1.40 A
1.41 C 1.42 60
Thermal conductivity is lower for air. Hence, the correct option is (B). Key Point (i) In case of solids thermal conductivity of
material is due to flow of free electrons and lattice vibration.
(ii) In case of liquids and gases, thermal conductivity is due to molecular wave.
(iii) Pure metals will have highest thermal conductivity from its alloys.
'puremetal it s alloyk k , iron steelk k ,
copper brassk k , ice water water vapour airk k k k
Maximum thermal conductivity is of diamond due to crystalline lattice. 2300 W/mKdiamondk
Key Point (i) The ratio of the change in pressure to the
volume compression is the Bulk Modulus.
22
3 3
N/m N/m/ m /mp
KV V
(ii) Thermal conductivity is ability of material to conduct heat through it. S.I. unit W/m K.
2
Wm K/m
Qk
dTA
dx
W/m K
(iii) Heat transfer coefficient (h) is used as a proportionality constant between heat flow and driving force for heat flow, i.e., temperature difference.
22
W W/m Km K
Qh
A T
Given : Wire diameter (d) = 20 mm Heat transfer coefficient 225 W/m Kh Thermal conductivity 0.5 W/mKk Critical radius of insulation,
0.5 0.020m25c
kr
h = 20 mm
or 20 10mm2 2d
r
Answers Conduction
Explanations Conduction
1.1 (B)
1.2 A-2, B-5, C-6, D-4
1.3 (B)
3.11GATE ACADEMY® Heat Transfer : Conduction
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Hence, thickness of insulation
c ct r r 20 10 10mm
Hence, the correct option is (B).
Key Point In case of plane wall the area
perpendicular to the direction of heat flow adding more insulation to a wall always decreases heat transfer. The thicker the insulation, the lower the heat transfer rate. This is due to the fact the outer surface have always the same area.
But in cylindrical and spherical coordinates, the addition of insulation also increases the outer surface, which decreases the convection resistance at the outer surface. Moreover, in some cases, a decrease in the convection resistance due to the increase in surface area can be more important than an increase in conduction resistance due to thicker insulation. As a result the total resistance may actually decrease resulting in increased heat flow.
The thickness upto which heat flow increases and after which heat flow decreases is termed as critical thickness. In the case of cylinders and spheres it is called critical radius. It can be derived the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h.
Two insulating materials of thermal conductivity k and 2k are available for lagging a pipe carrying a hot fluid here the critical radius for insulation must be
(i) For first material 1
( )c
kr
h
(ii) For second material 2
2( )c
kr
h
It is apparent that 2 1
( ) 2c cr r
Hence, the first material is to be used inside and second material is to be used for outside as it has bigger critical radius.
Hence, the correct option is (B).
For heat flow over the same thickness, the temperature drop across the material will be maximum for that material which is having more resistance to heat flow i.e., value of thermal conductivity will be less, from Fourier equation we can infer as follows :
Fourier equation,
dTQ kA
dx
QdxkdT
A
kdT Constant or 1dT
k
Hence, the correct option is (C).
Key Point
glass wool refractory bricks steel copperk k k k
Whichever the material is having lowest thermal conductivity the corresponding material has highest temperature drop.
r
rc
tc
1.4 (B)
1.5 (C)
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From Fourier law, we can infer as follows :
dTQ k
dx as Q constant.
1dT
dx k
1 1 2 2
Constant Constant,dT dT
dx k dx k
1 21 2
ordT dTk k
dx dx
Hence, the correct option is (A).
Key Point (i) Here, k is called thermal conductivity of
Material that measures the ability of material to conduct heat. It is a property that depends on temperature but not on too extent. For all gases, k increase with temperature due to collision as result vibration increase.
(ii) On the other side we can infer conductivity as opposite of resistivity, the more resistance, temperature drop will be more.
Given : Wire diameter ( ) 1 mmd Thermal conductivity of enamel point ( ) 0.1W/mKk
Heat transfer coefficient 2( ) 100 W/m Kh
Critical radius,
0.1 0.001 m 1mm100c
kr
h
Critical thickness of enamel point
11 0.5mm2c ct r r
Hence, the correct option is (B).
Out of the given substances pure aluminum has high conductivity and steam has low conductivity. Aluminum - 250 W/mK, Iron - 80 W/mK, Liquid water - 0.7 W/mK, Water vapour - 0.016 W/mK Hence, the correct option is (C).
Given : Temperature of water 0( ) 48 CwT
Temperature of glass plate 0( ) 40 CgT
Thermal conductivity of water( ) 0.6 W/mKwk
Thermal conductivity of glass ( ) 1.2 W/mKgk
Spatial gradient of temperature of water-glass
interface, 41 10 K/mw
dT
dy
Heat generation (flux),
w g
w g
dT dTq k k
dy dy
w
gg w
kdT dT
dy k dy
4 40.6 (1 10 ) 0.5 10 K/m1.2
g
dT
dy
Hence, the correct option is (C)
Key Point The temperature gradient in the glass interface must be lesser than water as here thermal conductivity is higher, that suggest the resistance will be lesser in glass.
r
rc
tc
1.6 (A)
1.7 (B)
1.8 (C)
1.9 (C)
3.13GATE ACADEMY® Heat Transfer : Conduction
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At steady state condition, heat will pass through water-glass interface from either side i.e. from water to glass and vice versa.
At interface ( 0)y , following the no slip condition, water will not move. Hence at 0y , heat transfer will be due to conduction in water and then the same amount of heat will be convected into water.
0
( )water water glass
y
dTk A h A T T
dy
40.6 1 10 8h 2750 W/m Kh Hence, the correct option is (D).
One dimensional unsteady state heat transfer equation in spherical coordinate system is
22
Internal heat Unsteady conduction heatgeneration term termtransfer term
1 1T q Tr
r r r k t
Hence, the correct option is (B).
Key Point The one dimensional time dependent heat conduction equation can be written more compactly as in a simple manner as,
1 nn
T q c Tr
r r r k k t
Where, n = 0, for rectangular section n = 1, for cylindrical section n = 2, for spherical section.
k
c
Where, is called thermal diffusivity and ability to transport heat through conduction.
Given : Temperature difference between innermost and outermost surface 0
1 2( ) 600 CT T Thermal conductivity of stainless steel ( ) 19 W/mKsk Thermal conductivity of asbestos ( ) 0.2 W/mKak
Internal diameter of tube ( ) 2cmid
Outer diameter of tube 0( ) 5cmd
Thickness of insulation ( ) 3cmit Length of tube (L) = 1 m
Insulation diameter of asbestos, 2 5 2 3 11 cmc o id d t
0 5ln ln2
2 2 19 1i
steels
d
dR
k L
0.00767 K/WsteelR
0
11ln ln5
2 2 0.2 1
c
asbestosa
d
dR
k L
0.6274asbestosR K/W
WaterGlass
(Stagnation Layer)048 C
040 C
y
ri
tc
r0
rc
L = 1 m
Insulation
1T aT
2T
steelR asbestosR
1.10 (D)
1.11 (B)
1.12 (C)
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1 2 600 6000.00767 0.6274th s a
T TQ
R R R
944.77 944.72 W/mQ Hence, the correct option is (C).
1h c
s
T Tq
R
and 2
h c
s c
T Tq
R R
Because s c sR R R 1 2q q
1Resistance
Q
Resistance is inversely proportional to thermal conductivity. Hence it is fact that conductivity of steel is less than conductivity of copper. s cR R
h i i cT T T T
Q Q
h i i cT T T T 2h c iT T T
2
h ci
T TT
Hence, the correct option is (D).
For one dimensional heat conduction, we have
2
2
1T T
x t
Since, (thermal diffucsivity) is constant as it a property, we have
2
2
T T
t x
Hence, the correct option is (D).
Given : Thermal conductivity of 1st slab
1( ) 0.02 W/mKk Thermal conductivity of 2nd slab
2( ) 0.10 W/mKk
Thermal conductivity of 3rd slab
3( ) 0.04 W/mKk
Length of 1st slab 1( ) 0.5mL
Length of 2nd and 3rd slab 2 3( ) ( ) 0.25mL L Assumption : (i) Heat transfer is steady since there is no
indication of change with time. (ii) Heat transfer can be approximated as
being one-dimensional since it is predominantly in the x-direction.
(iii) Thermal conductivities are constant. (iv) Heat transfer by radiation is negligible. This given composite slab can redrawn in the form of thermal circuit.
Assuming unit width of slab,
11
1
LR
k A
0.50.02 1 1
25K/W
22
2
LR
k A
0.250.1 0.5 1
5K/W
33
3
LR
k A
0.250.04 0.5 1
12.5K/W
Overall thermal resistance,
2 31
2 3th
R RR R
R R
5 12.525(5 12.5)
28.6K/WthR Hence, the correct option is (C).
Key Point There is no variation in the horizontal (x-direction). So, we consider a unit depth and unit height portion of the slab since it is representative of the entire wall. Assuming any cross- section of the slab normal to the x-direction to be isothermal.
2R
A Bq
1R
3R
1.13 (D)
1.14 (D)
1.15 (C)
3.15GATE ACADEMY® Heat Transfer : Conduction
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Given : 1 2
2i
T TT
Thermal circuit,
At steady state, 1 2Q Q
1 2 1 21 2
1 2
2 22
T T T TT T
Qb b
k A k A
1 1 2 2 1 2( ) ( )2 1
k T T k T TQ
1 22k k
Hence, the correct option is (D).
Heat loss to surrounding in a circular pipe,
0
0
2 ( )ln[ / ]1 i
L TQ
r r
hr k
The addition of insulation to a cylindrical piece or spherical shell increase the conduction resistance of the insulation layer but decrease the convection resistance of the surface because of the increase in the outer surface area for convection. The heat transfer from the pipe may increase or decrease, depending on which effect dominates. Hence, the correct option is (A).
Given : Inner diameter of glass cylinder ( ) 0.03mid
Outer diameter of glass cylinder 0( ) 0.05md
Thermal conductivity of glass ( ) 1.05 W/mKk
This is case of radial heat transfer,
Resistance network for above figure is,
0ln
2i
th
r
rR
kL
0.05ln0.03
2 (1.05) 1
(Per unit length 1mL )
0.077 K/WthR per unit length
Hence, the correct option is (B).
Given : Diameter of nuclear reactor fuel rod
3( ) 50mm 50 10 md Convective heat transfer coefficient
2 3 2( ) 1kW/m K 1 10 W/m Kh Heat generation with in the fuel element
6 3( ) 50 10 W/mq
Ambient temperature 0( ) 75 CT
Under steady state conditions, by energy conversion. generated convection lossQ Q
1T 2T
1 2
2
T T�� � � � 2
b
k A1
2b
k A
Q Q
ri
r0
Hot fluid
1.0 m
iT 0T
�0ln /
2
ir rR
Lk�
�
Fuel element
Convective losssT T��
q�
1.16 (D)
1.17 (A)
1.18 (B)
1.19 (A)
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( ) ( )sq V hA T T
2 ( )4 sq d L hA T T
2 ( )( )4 sq d L h dL T T
( )4 s
dq h T T
6 33 050 10 50 10 1 10 ( 75 )
4 sT
0625 75 700 CsT Hence, the correct option is (A).
Given : Thickness of plate ( ) 20 mmt
Uniform heat generation 3( ) 80MW/mq
Temperature at the left face 01( ) 160 CT
Temperature at the right face 02( ) 120 CT
Thermal conductivity of the plate ( ) 200 W/mKk
..Method 1.. From steady state one dimensional heat conduction equation with heat generation
2
2 0d T q
dx k
2
2
d T q
dx k
Integrating both sides,
1dT q
x Cdx k
…(i)
Again integrating,
2
1 22qx
T C x Ck
…(ii)
From equation (ii), At, 0,x 0160 CT 2 160C
At, 20mm,x 0120 CT
6
21
80 10120 (0.02) 0.02 1602 200
C
1 2000C
For maximum temperature 0dT
dx
1qx
Ck
1C kx
q 6
2000 20080 10
5mm
Hence, the correct option is (C).
..Method 2.. Location of maximum temperature for steady state condition with internal heat generation when walls are maintain at different temperature.
2 1
max 2g
g
q LT Tkx
q L k
6
max 6
200 120 160 80 10 0.0280 10 0.02 2 200
x
3max 5 10 mx or 5 mm (from left face)
Hence, the correct option is (C).
..Method 3..
0160 C
0120 C
q�
x
1T
2T
x L�0x �
1T
2T
x L� � x L�0x �
2L
0.01mx � � 0.01mx � �
1.20 (C)
3.17GATE ACADEMY® Heat Transfer : Conduction
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max 2 1( )2 g
kx T T
q L
max 6
200 (120 160)2 80 10 0.01
x
3max 5 10 mx
max 5mmx (from mid place )
i.e., max 5 10 5 5mmx L (from left face).
Key Point In case of heat generation maximum temperature will occur where the heat is being generated. (i) If heat is being generated within the wall
at any location.
(a) If 1 2T T , x L (b) If 1 2T T , x L 1 2gQ Q Q
(c) maxx is location of maxT
At maxx , 0dT
dx
(i.e., slope of temperature profile is zero which represent maximum temperature).
(ii) If heat is being generated one side of the wall. Temperature will be maximum on that side of wall.
Here heat is being generated on left hand
face, hence 1T is the maximum temperature.
gQ Q 1 2
2gq LT T
kAL k
.
According to question, by using equation (ii),
2
1 22qx
T C x Ck
max 5( )x mmT
6280 10 (0.005)
400
0.005 2000 160 0
max 5mm( ) 165 CxT
Hence, the correct option is (B).
Given : For surface Q :
10T
x
K/m,
Thermal conductivity of body ( ) 0.1k W/mK For surface P :
0T
x
Heat is transferred only in y direction.
T
maxT
1T1Q 2Q
2T
x
L
Plane where heat isbeing generatedat a rate of Qg
1T
2T
L
gQQ
0160 C
0120 C
q�
x
0165 C
1.21 (B)
1.22 (D)
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For steady state condition,
P Q
T TkA kA
y x
( 2 )Q PA A
2T T
y x
2 10T
y
20T
y
K/m
Hence, the correct option is (D).
Key Point Features of Fourier law : Fourier law is valid for all matter solid,
liquid or gas. The vector expression indicating that
heat flow rate is normal to an isotherm and is in the direction of decreasing temperature.
It cannot be derived from first principle. It helps to define the transport property
‘k’.
Thermal conductivity, 0k k bT
Since, heat conduction is one dimensional
Heat flux, q dTk
A dx
As x increase, temperature in the wall increase and so the thermal conductivity increase, then
the value of dT
dx
must decreases to ensure
constant heat flux. Hence, the correct option is (D).
gq Heat generation per unit volume.
Steady state, one dimensional heat conduction equation,
2
2 0gqd T
dx k
2
2gqd T
dx k
Integrating both sides,
1gqdT
x Cdx k
Again integration both sides,
21 22
gqT x C x C
k …(i)
Now, applying the boundary condition on equation (i), At 0,x 0T then,
2 0C
At ,x L 0100 CT then,
2
11002gq L
C Lk
1100
2gq L
CL k
Surface
Surface
0,0 CQ
0,100 CP
y
x
0100 C
00 C
q
0x � x L�x
1.23 (D)
1.24 (B)
3.19GATE ACADEMY® Heat Transfer : Conduction
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So that, equation (i) becomes
2 100
2 2g gq x q L
T xk L k
This equation shows that temperature distribution in the wall will not be linear but parabolic. For maximum temperature,
0dT
dx
2 100 02 2
g gq x q L
k L k
1002
g
g
q Lkx
q L k
1002g
k Lx
q L
So the curve shows that maximum temperature will be more than that of 0100 C, which will be close to x L and heat will flow, both sides of this maximum temperature plane. Hence, the correct option is (B).
Let, iT Inner surface temperature of
cylindrical tube 0T Outer surface temperature of
cylindrical tube Using electrical analogy,
0i
th th
T T TQ
R R
…(i)
From Fourier law for one dimensional (radial) heat transfer through pipe,
Consider an elementary cylinder of radius ‘r’ and thickness dr. Area, 2A rL (where, L is length of tube) From Fourier law, (under steady state condition),
2dT dTQ kA k rL
dr dr
0 0
2i i
r T
r T
Q drdT
kLr
0 0ln2 i i
r Tr T
Qr T
kL
0 0(ln ln ) ( )2 i i
Qr r T T
kL
0
01 ln2
i
i
T TQ
r
kL r
…(ii)
Comparing equation (i) and (ii),
01 ln2th
i
rR
kL r
Hence, the correct option is (C).
3
v
A
nV ck
N
Where, k = Thermal conductivity V = Mean particle speed = Mean free path vc = Molar heat capacity
AN = Avogadro's number
n = Particles per unit volume Gases transfer heat by direct collisions between molecules. As the temperature increases, the thermal conductivity increases due to increase in speed, movement and collisions in the molecules. From the above expression, by increasing mean particle speed, the thermal conductivity increases. Hence, the correct option is (A).
iTQ
oTthR
0r
ir
r
dr
1.25 (C)
1.26 (A)
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Given : Thickness of wall ( ) 30mmL Thermal conductivity of wall ( ) 15W/mKk
Heat flux at 0x 5 2( ") 1 10 W/mq Convective heat transfer coefficient
2( ) 250 W/m Kh
Temperature of surrounding 0( ) 25 CT
For the given problem, the network of thermal resistance involves two resistances in series as shown in figure,
Conductive thermal resistance,
cond
LR
kA
Convective thermal resistance,
1convR
hA
Rate of heat transfer,
th
TQ
R
1 1
1cond conv
T T T TQ
LR RkA hA
Heat flux,
1"1
T Tq
Lk h
5 1 251 10 0.03 115 250
T
5 1 251 100.006T
5 01 (1 10 0.006) 25 625 CT
Hence, the temperature at 0x is 6250C.
Given : Thickness of 1st steel slab 1( ) 20mml
Thickness of 2nd steel slab 2( ) 20mml
Thickness of P material 3( ) 1mml
Heat flux 2( ) 10kW/mq
Thermal conductivity of 1st and 2nd steel slab
1 2( ) ( ) 10 W/mKk k
Temperature of 1st steel slab 1( ) 500KT
Temperature of 2nd steel slab 3( ) 360KT
As per steady state heat conduction equation,
According to question applying steady state heat conduction equation. Heat flux is given by,
1 2
31 2
1 2 3
" T Tq
ll l
k k k
3 3
2
500 3601000020 10 0.001 20 10
10 10k
0�x
x
025 C� �T
2T
1T
5 2" 1 10 W/m� �q
T1Rconductive RconvectiveT2 T� = 25 C
0
0.02 m 0.02 mm
2 360K�T2" 10 kW/m�q
1 500 K�T
2k
PSteelslab
Steelslab
1k 3k
1.27 625
1.28 0.1
3.21GATE ACADEMY® Heat Transfer : Conduction
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2
14010000 0.0010.002 0.002k
2 0.1W/mKk
Hence, the thermal conductivity of material P is 0.1 W/mK.
Given : Thermal conductivity of 1st section 1( )k k
Thermal conductivity of 2nd section 2( ) 2k k
Temperature of left face of 1st section
1( ) 600KT
Temperature of right face of 2nd section
2( ) 300KT
At steady state, heat transfer is given by, Fourier’s equation of heat conduction, Q = – kA(dT/dx)
For 1st section,
1 1dT
Q k Adx
…(i)
For 2nd section,
2 2dT
Q k Adx
…(ii)
Now, equating equations (i) and (ii) 1 2Q Q
1 2dT dT
k A k Adx dx
1 2 21
( ) ( )i iT T k A T Tk A
l l
1 2( ) 2 ( )i ikA T T kA T T
l l
1 22 2i iT T T T
1 22 2 i iT T T T
1 22 3 iT T T
1 22 600 2 300 400K3 3i
T TT
Hence, the interface temperature iT of the composite wall is 400 K.
Given : Thermal conductivity ( ) 1.15W/mKk
Inner surface temperature 0( ) 1100 CiT
Outer surface temperature 00( ) 350 CT
Heat flux 2( ) 2500 W/mq
Fourier’s equation of heat conduction, Q = – kA(dT/dx)
0ik T Tq
L
1.15 1100 3502500
L
Solving, we get 0.345mL Hence, the design thickness of the wall to maintain a steady heat flux of 22500 W/mshould be 0.345 m.
l l
300 K600 K iT
k 2k
Heat flow
Q2Q1
1 2
0
0 350 CT �
01100 CiT �
22500W/mq �
L
1.29 400
1.30 0.345
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Given : Diameter of pipe ( ) 4 cmd Convective heat transfer coefficient inside pipe
2( ) 15 W/m Kih
Convective heat transfer coefficient outside foam insulation 2
0 2 W/m Kh
Thermal conductivity of pipe ( ) 15 W/mKpipek
Thermal conductivity of foam insulation ( ) 0.1 W/mKfoamk
According to question it is given that the pipe is insulated by foam from outer side. So, we find the critical radius of insulation.
0
0.1 0.05m2
foamc
kr
h
Critical radius of insulation 5cmcr .
Hence, the critical radius of insulation is 5 cm.
Key Point In finding critical insulation the thermal conductivity of the insulation material is taken and hence, here there is no use of the thermal conductivity of the pipe.
Given : Diameter of electrical conductor ( ) 10mmd ( ) 5 mmr Thickness of insulation ( ) 2mmt Thermal conductivity of insulation ( ) 0.08W/mKik
Convective coefficient of insulation surface 2( ) 10 W/m Kih
Critical radius, 0.08 0.008m10
ic
i
kr
h
8 mmcr
Thickness of critical insulation is, 8 5 3mmc ct r r Heat flow for various radius is shown in above figure. Heat loss will increase up to thickness 3 mm of insulation then it will decreases. Hence, the correct option is (C).
Given : Thermal conductivity of brick wall ( ) 0.9 W/mKk Thickness of brick wall ( ) 0.18mL Outside (surrounding) temperature of air
0( ) 5 CT
Temperature of room 0( ) 27 CiT Convective heat transfer coefficient
2( ) 20 W/m Kh
Total thermal resistance,
1 11
thLR
h k
1 1 41 0.1820 0.9
thR
Q
Increase
Decrease
2 mm
3 mm
Thicknessof insulation
220W/m Kh �027 CiT �
0.18mL �
05 CT� � �Wall
0.9W/mkk �
1.31 5
1.32 (C)
1.33 (C)
3.23GATE ACADEMY® Heat Transfer : Conduction
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1 W/K4thR
Heat loss inside the room is,
227 ( 5) 128W/m14
th
TQ
R
Hence, the correct option is (C).
Given : Radius of uranium fuel rod ( ) 5mm 0.005 mr Heat generation in nuclear reactor
7 3( ) 4 10 W/mgq
Convective heat transfer coefficient 2( ) 1000 W/m Kh
Surrounding temperature 0( ) 25 CT
For steady state conduction, gen convQ Q
gq V hA T
2 (2 )gq r L h rL T
2gq r h T
74 10 0.005 2 1000( )sT T
74 10 0.005( ) 100
2 1000sT T
100sT T
0100 25 125 CsT
But in this question temperature asked in Kelvin so, 125 273 398KsT
Hence, the correct option is (B).
Given : Outer radius of plastic sleeve 0( ) 1mmr
Electric current carrying wire radius( ) 0.5mmr
Thermal conductivity of plastic ( ) 0.15W/mKk Convective heat transfer coefficient
2( ) 25W/m Kh According to question it is said that the wire is insulated by a plastic sleeve which shown in figure below,
Critical radius, 30.15 6 10 m25c
kr
h
6mmcr Here, we can see that the critical radius is greater than the radius of coating 0( )cr r . Therefore, addition of plastic sleeve will increase the heat transfer. Hence, the correct option is (A).
L Length, k Thermal conductivity
Thermal resistance given by 2 1ln( / )2r r
kL
Hence, the correct option is (A).
Given : Temperature at face 1 0
1( ) 130 CT
Temperature at face 3 03( ) 30 CT
Thermal conductivity of slab A ( ) 20 W/mKAk
r
r0
Wire
Plastic
h = 25 W/m K2
k=
0.15 W/mK
k
1r
L
r2r2
1.34 (B)
1.35 (A)
1.36 (A)
1.37 67.5
3.24 Topic Wise GATE Solutions [ME] GATE ACADEMY®
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Thermal conductivity of slab B ( ) 100 W/mKBk Length of slab A ( ) 0.1mAL Length of slab B ( ) 0.3mBL
The thermal resistance network for the representative section of slab becomes as shown in figure. Then the individual thermal resistance is,
1
0.120
A
A
LR
k and 2
0.3100
B
B
LR
k
For steady one-dimensional heat conduction,
1 3 1 2
1 2 1
T T T TQ
R R R
2130130 300.1 0.3 0.120 100 20
TQ
21301000.8 0.1100 20
T
2130 62.5T
02 130 62.5 67.5 CT
Hence, the interface temperature 2T is 67.5 0C .
Given : Diameter of solid cylinder rod ( ) 10 mmD Heat generated in rod 7 3( ) 4 10 W/mgq Thermal conductivity of the rod ( ) 25W/mKk
Temperature distribution in a cylindrical rod with uniform heat generation under steady state is given by
2 2( )
4g
r s
q R rT T
k
At center 0r and 0rT T
2
0 4g
s
q RT T
k
And radius of solid cylinder,
10 5 mm2 2D
R
27 3
00
4 10 5 10 10 C4 25sT T
Hence, the temperature difference between the centre and the surface of the rod is 010 C .
Key Point Under steady state one dimensional heat conduction in cylinder is derived below,
gq rTr
r r k
…(i)
Integrating above equation, we get
2
12gq rT
r cr k
1
2gq r cT
r k r
At 00,r T T and 0T
r
1 0c
2
gq rT
r k
…(ii)
Again integrating equation (ii),
2
24gq r
T ck
…(iii)
A B
1 2 3
0
1 130 CT � 0
3 30 CT �20
W/m.K
Ak �100 W/m.KBk �
Q Q Q
0.1 m 0.3 m
T1 = 130 C0 R1 R2T2 T3 = 30 C
0
Q
R
Ts
T0
1.38 10
3.25GATE ACADEMY® Heat Transfer : Conduction
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At , sr R T T
2
24g
s
q RT c
k
2
2 4g
s
q Rc T
k
Putting the value of 2c in equation (iii),
2 2
4 4g g
s
q r q RT T
k k
2 2
4 4g g
s
q R q rT T
k k
2 2( )4
gs
qT T R r
k …(iv)
In case of solid cylinder value of r is zero then the equation (iv) becomes,
2
4g
s
q RT T
k
Given : Thickness of slab ( ) 20cmL
Thermal conductivity of slab ( ) 2.5W/mKk
Convective heat transfer coefficient 2( ) 10 W/m Kh
Ambient temperature 0( ) 45 CT
Surface temperature 0( ) 25 CQT
Considering steady state condition,
Thermal resistance circuit for the given problem as,
1 110convR
hA A
0.22.5cond
LR
kA A
For steady state one dimensional heat generation,
conv cond
T TQ
R R
1 0.210 2.5
P QPT TT T
Q
A A
45 251 0.2
10 2.5
P PT TQ
45 250.1 0.08
P PT TQ
0.08(45 ) 0.1( 25)P PT T
033.88 CPT
Hence, the steady state temperature pT of the
side which is exposed to the fluid is 033.88 C .
1 2th th thR R R
1 2
2
e
L L L
k k k
1 2
1 2 1 2
2 1 1
e
k k
k k k k k
1 2
1 2
2e
k kk
k k
Hence, the correct option is (A).
L = 20 cm
TQ = 25 C0h = 10 W/m K
2
0= 45 CT�
TP
k = 2.5 W/mK
045 CT� � PT
convR condR
025 CQT �
k1 k2
L L
T�
dQ
ke
2L
T�
Q
1.39 33.88
1.40 (A)
3.26 Topic Wise GATE Solutions [ME] GATE ACADEMY®
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B Oranch ffice : Raipur : 79743-90037, Bhopal : 83198-88401
Fourier law, assumptions are, 1. Steady state, one dimensioned. 2. No internal heat generation. 3. Thermal conductivity = Constant
dTQ kA
dx
Given, area is varying linearly. A a bx (a and b are constant) Assuming, 0x , 0T and x L , T T
So, ( )Q dT
K a bx dx
0 0( )L TQ
dx dTK a bx
ln( )Qa bL T
bK
So, temperature is varying logarithmic. Hence, the correct option is (C).
According to question,
10000 ( )gE L A
10000 1 1 10000 WgE
and, 1 11( ) conv condhA T T q q
1
100 (100 30) condq
1
7000 Wcondq
Again, 2 2( )convq hA T T 2100 1 ( 30)T
2 22100( 30)conv condq T q
At steady state, g outE E
1 2g cond condE q q
210000 7000 100( 30)T
02
3000 30 60 C100
T
Hence, the right extreme face temperature 2T is 600C.
T1 = 100 C0
h = 100 W/m K2
T2 = ?
T� = 30
qcond 1 qcond 2
T� = 30
h = 100 W/m K2
1 m 1 m 1 m
5 W/m Kk � 5 W/m Kk �
1.41 (C)
1.42 60